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Question Number 39827 by Rio Mike last updated on 11/Jul/18
if f(x) = 3x^3 + px^2 + 4x − 8 and (x − 1) is a factor of f(x). a) find the value of p. with these value of p b) solve the equation f(x) = 0. if α and β are roots of f(x), c) find α + β and αβ d) Evaluate α^2 + β^2 hence α − β
$${if}\:{f}\left({x}\right)\:=\:\mathrm{3}{x}^{\mathrm{3}} \:+\:{px}^{\mathrm{2}} \:+\:\mathrm{4}{x}\:−\:\mathrm{8} \\ $$$${and}\:\left({x}\:−\:\mathrm{1}\right)\:{is}\:{a}\:{factor}\:{of}\: \\ $$$${f}\left({x}\right). \\ $$$$\left.{a}\right)\:{find}\:{the}\:{value}\:{of}\:{p}. \\ $$$${with}\:{these}\:{value}\:{of}\:{p} \\ $$$$\left.{b}\right)\:{solve}\:{the}\:{equation}\:{f}\left({x}\right)\:=\:\mathrm{0}. \\ $$$${if}\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\: \\ $$$${f}\left({x}\right),\: \\ $$$$\left.{c}\right)\:{find}\:\alpha\:+\:\beta\:{and}\:\alpha\beta \\ $$$$\left.{d}\right)\:{Evaluate}\:\alpha^{\mathrm{2}} \:+\:\beta^{\mathrm{2}} \:{hence}\:\alpha\:−\:\beta \\ $$$$ \\ $$
Answered by MJS last updated on 11/Jul/18
3(x−1)(x−α)(x−β)=3x^3 +px^2 +4x−8 3x^3 −3(α+β+1)x^2 +3(α+αβ+β)x−3αβ= =3x^3 +px^2 +4x−8 ⇒ p=−3(α+β+1) 3(α+αβ+β)=4 3αβ=8 ⇒ α=(8/(3β)) 3((8/(3β))+(8/(3β))β+β)=4 ⇒ β^2 +(4/3)β+(8/3)=0 ⇒ ⇒ β=−(2/3)±((2(√5))/3)i ⇒ α=−(2/3)∓((2(√5))/3)i ⇒ p=1 α+β=−(4/3) αβ=(8/3) α^2 +β^2 =−((32)/9) α−β=±((4(√5))/3)i
$$\mathrm{3}\left({x}−\mathrm{1}\right)\left({x}−\alpha\right)\left({x}−\beta\right)=\mathrm{3}{x}^{\mathrm{3}} +{px}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{8} \\ $$$$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{3}\left(\alpha+\beta+\mathrm{1}\right){x}^{\mathrm{2}} +\mathrm{3}\left(\alpha+\alpha\beta+\beta\right){x}−\mathrm{3}\alpha\beta= \\ $$$$=\mathrm{3}{x}^{\mathrm{3}} +{px}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{8} \\ $$$$\Rightarrow \\ $$$${p}=−\mathrm{3}\left(\alpha+\beta+\mathrm{1}\right) \\ $$$$\mathrm{3}\left(\alpha+\alpha\beta+\beta\right)=\mathrm{4} \\ $$$$\mathrm{3}\alpha\beta=\mathrm{8}\:\Rightarrow\:\alpha=\frac{\mathrm{8}}{\mathrm{3}\beta} \\ $$$$\mathrm{3}\left(\frac{\mathrm{8}}{\mathrm{3}\beta}+\frac{\mathrm{8}}{\mathrm{3}\beta}\beta+\beta\right)=\mathrm{4}\:\Rightarrow\:\beta^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}\beta+\frac{\mathrm{8}}{\mathrm{3}}=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\beta=−\frac{\mathrm{2}}{\mathrm{3}}\pm\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{3}}\mathrm{i} \\ $$$$\Rightarrow\:\alpha=−\frac{\mathrm{2}}{\mathrm{3}}\mp\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{3}}\mathrm{i} \\ $$$$\Rightarrow\:{p}=\mathrm{1} \\ $$$$\alpha+\beta=−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\alpha\beta=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =−\frac{\mathrm{32}}{\mathrm{9}} \\ $$$$\alpha−\beta=\pm\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{3}}\mathrm{i} \\ $$
Commented by Rio Mike last updated on 12/Jul/18
the cubic expression is of the form y = ax^(3 ) + bx^2 + cx + d so what is α+β^ =? and αβ=?
$${the}\:{cubic}\:{expression}\:{is}\:{of}\: \\ $$$${the}\:{form}\: \\ $$$${y}\:=\:{ax}^{\mathrm{3}\:} +\:{bx}^{\mathrm{2}} \:+\:{cx}\:+\:{d} \\ $$$${so}\:{what}\:{is}\:\alpha+\overset{} {\beta}=? \\ $$$${and}\:\alpha\beta=? \\ $$
Commented by MJS last updated on 12/Jul/18
α, β are the roots besides 1 which is given in this case. ax^3 +bx^2 +cx+d=a(x−α)(x−β)(x−γ)= =a(x^3 −(α+β+γ)x^2 +(αβ+αγ+βγ)x−αβγ) in our case γ=1 leads to a(x^3 −(α+β+1)x^2 +(α+αβ+β)x−αβ)
$$\alpha,\:\beta\:\mathrm{are}\:\:\mathrm{the}\:\mathrm{roots}\:\mathrm{besides}\:\mathrm{1}\:\mathrm{which}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in} \\ $$$$\mathrm{this}\:\mathrm{case}. \\ $$$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}={a}\left({x}−\alpha\right)\left({x}−\beta\right)\left({x}−\gamma\right)= \\ $$$$={a}\left({x}^{\mathrm{3}} −\left(\alpha+\beta+\gamma\right){x}^{\mathrm{2}} +\left(\alpha\beta+\alpha\gamma+\beta\gamma\right){x}−\alpha\beta\gamma\right) \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:\gamma=\mathrm{1}\:\mathrm{leads}\:\mathrm{to} \\ $$$${a}\left({x}^{\mathrm{3}} −\left(\alpha+\beta+\mathrm{1}\right){x}^{\mathrm{2}} +\left(\alpha+\alpha\beta+\beta\right){x}−\alpha\beta\right) \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jul/18
AnOther approach If f(x) is divided by x−1 then the quotient Q(x) and remainder R can be determined by synthetic division: (((1) 3),( p),( 4),( −8)),(,( 3),(p+3),( p+7)),(( 3),(p+3),(p+7),(∣p−1^(−) )) ) ∵ x−1 is factor of f(x) ∴R=p−1=0⇒p=1 (a) Q(x)=3x^2 +(p+3)x+(p+7) =3x^2 +(1+3)x+(1+7) =3x^2 +4x+8 f(x)=(x−1).Q(x)+R [Relation between divedend,divisor & remainder] =(x−1).Q(x)+0 =(x−1)(3x^2 +4x+8) (b) f(x)=0⇒x−1=0 ∣ 3x^2 +4x+8=0 x=1 ∣ x=((−4±(√(16−96)))/6) ∣ x=((−2±2i(√5))/3) Let α & β are the roots of 3x^2 +4x+8=0 α+β=−(4/3) & αβ=(8/3) α−β=((−2+2i(√5))/3)−((−2−2i(√5))/3) =((4(√5))/3)i α^2 +β^2 =(α+β)^2 −2αβ=(−(4/3))^2 −2((8/3)) =((16)/9)−((16)/3)=((16−48)/9)=−((32)/9)
$$\mathrm{AnOther}\:\mathrm{approach} \\ $$$$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{x}−\mathrm{1}\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{quotient}\:\mathrm{Q}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{remainder}\:\mathrm{R}\:\mathrm{can} \\ $$$$\mathrm{be}\:\mathrm{determined}\:\mathrm{by}\:\mathrm{synthetic}\:\mathrm{division}: \\ $$$$\begin{pmatrix}{\left.\mathrm{1}\right)\:\mathrm{3}}&{\:\:\:\mathrm{p}}&{\:\:\:\mathrm{4}}&{\:\:−\mathrm{8}}\\{}&{\:\:\:\mathrm{3}}&{\mathrm{p}+\mathrm{3}}&{\:\:\mathrm{p}+\mathrm{7}}\\{\:\:\:\:\:\:\mathrm{3}}&{\mathrm{p}+\mathrm{3}}&{\mathrm{p}+\mathrm{7}}&{\mid\overline {\mathrm{p}−\mathrm{1}}}\end{pmatrix} \\ $$$$\because\:\mathrm{x}−\mathrm{1}\:\mathrm{is}\:\mathrm{factor}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\therefore\mathrm{R}=\mathrm{p}−\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{p}=\mathrm{1}\:\:\:\:\left(\mathrm{a}\right) \\ $$$$\mathrm{Q}\left(\mathrm{x}\right)=\mathrm{3x}^{\mathrm{2}} +\left(\mathrm{p}+\mathrm{3}\right)\mathrm{x}+\left(\mathrm{p}+\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{3x}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{3}\right)\mathrm{x}+\left(\mathrm{1}+\mathrm{7}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{8} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{1}\right).\mathrm{Q}\left(\mathrm{x}\right)+\mathrm{R} \\ $$$$\:\:\left[\mathrm{Relation}\:\mathrm{between}\:\mathrm{divedend},\mathrm{divisor}\:\&\:\mathrm{remainder}\right] \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}−\mathrm{1}\right).\mathrm{Q}\left(\mathrm{x}\right)+\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:=\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{8}\right) \\ $$$$\left(\mathrm{b}\right)\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{x}−\mathrm{1}=\mathrm{0}\:\:\mid\:\:\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}\:\mid\:\mathrm{x}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{96}}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\mathrm{x}=\frac{−\mathrm{2}\pm\mathrm{2i}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\mathrm{Let}\:\alpha\:\&\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{3x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\:\alpha+\beta=−\frac{\mathrm{4}}{\mathrm{3}}\:\:\&\:\:\alpha\beta=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\:\:\:\:\alpha−\beta=\frac{−\mathrm{2}+\mathrm{2i}\sqrt{\mathrm{5}}}{\mathrm{3}}−\frac{−\mathrm{2}−\mathrm{2i}\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{3}}\mathrm{i} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta=\left(−\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{8}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{16}}{\mathrm{9}}−\frac{\mathrm{16}}{\mathrm{3}}=\frac{\mathrm{16}−\mathrm{48}}{\mathrm{9}}=−\frac{\mathrm{32}}{\mathrm{9}} \\ $$

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